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You are not logged in. #1 20081121 08:16:04
Daniel's Challenge ThreadRather than me setting exercises for others to do, I'd like people to give me some challenging questions It's a thread where people can share nice problems, but I'd like the benefit of seeing if I can get to the result myself too. #2 20081121 08:54:08
Re: Daniel's Challenge ThreadHi. The solution uses material you should be familiar with (because I’ve seen you discuss it on the forum). Let me know if you need a hint. #3 20081121 10:41:19
Re: Daniel's Challenge ThreadIt would be helpful if we could multiply by a number n such that 13n is 1 more than a mutliple of 29. The smallest such number is 9. Mutliplying by 9 gives: i.e. Thank you Jane Last edited by Daniel123 (20081121 10:44:17) #4 20081130 00:09:40
Re: Daniel's Challenge ThreadHeres a VERY VERY VERY easy sum: #5 20081130 00:11:49
Re: Daniel's Challenge ThreadWriting the information out algebraically: #6 20081212 12:25:16#7 20081225 12:41:50
Re: Daniel's Challenge ThreadWell the first part comes out from the sum of a geometric series: Induction proves the second pretty quickly: Obviously the first part could also be done (and more quickly) by induction, but I feel induction gets rid of the 'why' behind it. I'm sure you're going to show me a nicer way to do the second part Last edited by Daniel123 (20081225 23:00:29) #8 20081228 14:43:33
Re: Daniel's Challenge ThreadWell, you can do both parts in a nice way, actually. Next one? #9 20090208 23:38:10#10 20100929 06:05:47
Re: Daniel's Challenge Thread
All we hv to do is to prove that 9k^2  1 is divisible by 8. 